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Is gravity less on an airliner at cruise altitude?
Is the force of gravity less on an airliner at cruise speed and altitude? I'm not talking about a special reduced gravity flight with a parabolic flight path, just a typical long distance flight.
Seems to me there should be less as passengers are further away from the centre of the earth, and also possibly a tiny effect from the speed the plane as it travels round the curve of the earth (altitude remaining constant) but does anybody actually know how much less?

Welcome to Aviation.SE!  This is really more of a Physics.SE type of question, since the answer is purely in the realm of physics and not aviation per-se.  Yes, because you are farther from the center of the earth flying at 40,000' above sea level than you are on the ground, the effect of gravity will be minutely less.  But for the distances involved, we're talking imperceptably slight, so much so that in practical terms, it's all the same.  But not exactly the same. Physics guys will be a better source of the formulas, math, and numbers you're looking for.

I'm voting to close this question as off-topic because it belongs on Physics.SE instead -- not about aviation, within the scope defined in the help center (aviation.stackexchange.com/help).

@RalphJ Is there a way of moving the question to physics SE instead of me having to retype it?

Why would you retype it?  Computers have had "Copy-n-Paste" for decades.

I think the diamond moderators can do that.

Yes, @Ralph can "flag" the question for moderator attention and request migration to Physics SE.

I'd be more inclined to move it if we didn't already have 2 good answers. If y'all really want us to, though, go for it.

The force of gravity is less and time passes faster.

Yes, gravity is less when you are flying.

@egid [physics.SE] mod here: FWIW on Physics we'd want to see the asker of a question like this show some effort to work out the answer themselves, or at least research it a bit. Topic-wise, it's definitely within our scope, but if this question had been posted as-is over there, it probably would not have been received so well. Just in case you do consider migrating it. (Also, there is some ambiguity in how the question is posed, as shown by the discrepancy between sweber's and the other answers.)

@DavidZ Makes sense.

There is less gravitational force, but by how much? An insignificant amount.
The gravitational force of attraction between two objects is given by,
$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,
where,
$G$ is the graviational constant,
$R$ is the distance between the object's centers, and
$m_{1}$ and $m_{2}$ are the masses of the objects.
Instead of finding the variation in force between the aircraft and earth, it would be be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)
We have, on earth's surface,
$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$
where,
$m_{\mathrm e}$ is the mass of the earth, and
$R_{\mathrm e}$ is the radius of the earth.
For the aircraft at an altitude $h$ above the surface of the earth, this becomes,
$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$
Taking ratio, we get,
$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$
Plugging in numbers, we get, for an airliner cruising at 12 km,
$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,
or about 0.37 % less compared to the sea level value. This is quite small and would not be noticable to all but the sensitive instruments.

That's over a ton for a fully loaded A380 :)

But a ton when you weigh 590 tons, is still pretty insignificant...

Most of those $R$'s should be $R^2$ instead.

The difference is big enough to be considered in A/C performance calculations, bigger than other effects that manufacturers spend significant amount of money (like parasitic drag of specific areas). However, is a data, a you can not do a lot to use it... bigger is the effect of air density. Relevant, but only for complex calculations.

does this change the stall speed in an accellerated stall?!

That depends on the direction, IE East/West. It could add perceived weight just as easily as remove it (by actually reducing it's overall rotational speed, up to the speed of the Earth's rotation at which point it would become negative again).

I'm afraid you are in fact wrong. You are trying to use a "fixed" Newtonian frame of reference, when in fact the correct frame of reference is to use the local frame of reference of the atmosphere, which we can approximate to a fluid that rotates around the earth once every 23.9 hours (a sidereal day).

My bathroom scales can tell me when my weight varies by 0.37%. I won't feel the difference, but when you say "sensitive instruments", it doesn't take special equipment to detect it.

for a Lockheed SR-71 Blackbird it is about 1.53% (according to my calculations)

@SteveJessop: Please take your bathroom scales on your next flight and perform the test in the lav.  I look forward to seeing your results.

The $\frac{g_h}{g}$ ratio is wrong. While $g$ is called "gravitational acceleration", it is actually defined as the total weight acceleration in the Earth frame of reference, i.e. including the centrifugal force due to own rotation of Earth. And since the centrifugal force increases with altitude, $g$ decreases faster than by that equation. Further complicated by the fact that aircraft speeds are significant compared to Earth rotation, so flying east or west does make a difference.

@Antzi a ton is a unit of mass. Mass is constant regardless of gravity. 1 ton at sea level = 1 ton at 35,000 feet = 1 ton on the moon

Gravity itself
@aeronalias is absolutely right.
Given the gravitational acceleration of $g=9.81m/s^2$ on the ground, a perfect spherical earth of radius $R_E=6370km$ with homogenous (at least: radially symmetric) density, one can calculate the gravitational acceleration at an altitude of $h=12km$ by
$$g(h)=g\cdot\frac{R_E^2}{(h+R_E)^2}= 9.773 \rm{m}/s^2$$
Expressed in terms of $g$, the difference is
$$g_\rm{diff} = 0.0368565736 m/s^2 = 0.003757g$$
Centrifugal forces
The question also asks for the centrifugal effect on the aircraft as it travels round the curve of the earth, which has not yet been answered yet. The effect is considered small, but compared to the effect on gravity itself, it isn't always.
I got some heavy objections on my answer and I have to admit, I really don't see their point. Therefore, I've edited this section and hope this helps.
In general, an object moving on a circular path experiences a centrifugal acceleration, pointing away from the center of the circle:
$$a_c=\omega^2r=\frac{v^2}{r}$$
$\omega=\frac{\alpha}{t}$ is the angular speed, i.e. the angle $\alpha$ (in radians) the object travels in a given time $t$ (in seconds).
Now let's consider a "perfect" Earth as described above, plus no wind.
A balloon hovering stationary over a point at the equator at 12km altitude will do one revolution ($\alpha=2\pi[=360°]$) in 24 hours. So it is $\omega=\frac{2\pi}{24\cdot60\cdot60s}$. Together with $r=R_e+h$, one gets for the balloon:
$$a_{cb}=0.03374061 m/s² = 0.0034394098 g$$
The circumference of the circle the balloon flies is $2\pi(R_e+h)=40099km$
Now consider an aircraft flying east along the equator at the same altitude at 250m/s (900km/h, 485kt) with respect to the surrounding air. (Keep in mind: no wind). In 24h, this aircraft travels a distance of 21600km, or 0.539 of the circumference. This means the aircraft does 1.539 revolutions of the circle in 24h, which means its angular speed is  $\omega=1.539\cdot\frac{2\pi}{24\cdot60\cdot60s}$.
Thus, the centrifugal force on the aircraft flying east is
$$a_\rm{ce} = 0.0799053814 m/s^2  =  0.0081452988 g$$
The same way, one can calculate what happens when the aircraft flies west:
$\omega=(1-0.539)\cdot\frac{2\pi}{24\cdot60\cdot60s}$
$$a_\rm{cw} = 0.0071833292 m/s^2   =   0.0007322456 g$$
Comparison
Let's write the values together to compare them. I've also added how much lighter a 100kg (220lb) person would feel due to the effects:
                                             | "weight loss"
g_diff = 0.0368565736 m/s²  = 0.003757 g     | 376gram (0.829lb)
a_cb   = 0.03374061   m/s²  = 0.0034394098 g | 344gram (0.758lb)
a_ce   = 0.0799053814 m/s²  = 0.0081452988 g | 815gram (1.797lb)
a_cw   = 0.0071833292 m/s²  = 0.0007322456 g |  73gram (0.161lb)

Note: The 100kg is what a scale at the North Pole (i.e. without any centrifugal effect) shows. The person already feels 344g lighter on the ground at the equator. The balloon doesn't change this (much).
But moving east/west has a larger effect on the weight than gravity alone. A person flying west feels even heavier than on ground!
Maybe another table, showing the weight of the person:
                                             kg      lb
1. Man at north pole                       100.00  220.46
2. Man at equator                           99.66  219.70
3. Man at equator, in balloon               99.28  218.88
4. Man at equator, in aircraft flying east  98.81  217.84
5. Man at equator, in aircraft flying west  99.55  219.47 <- More than 3.

The numbers shown are only valid at the equator and for flights east / west. In other cases, it becomes a little more complex.

EDIT: Being curious on how this depends on latitude, I created this plot about the absolute acceleration an aircraft experiences.

The radius in the equation of the centrifugal force is the distance of the aircraft to the axis of the Earth. It is clear that it decreases when moving away from the equator, and so does the acceleration.
The speed of the aircraft flying west will cancel out the speed of the earth at about 57° N / S, i.e. there is no centrifugal force. At larger latitude, the aircraft will fly in the opposite direction around the axis of the earth, building up a centrifugal force again.
Near the poles, both aircraft become  centrifuges (theoretically). E.g. flying a circle of 500m radius gives an acceleration of 12.7g. This is why the data rises to infinity there.
(When doing the math, one has to keep in mind that gravity always points to the center of the earth, while the centrifugal force points away from the axis. You can't just add them)

I don't understand your argument. A resting observer in space sees that the earth has a tangential speed of 464m/s at the equator. An aircraft flying east with 250m/s has a total tangential speed of 714m/s, and an aircraft flying west has a total tangential speed of 214m/s. What's wrong with this? (a tangential speed of 250m/s at 12km does not equal 250m/s on ground, but this effect is really negligible)

The question asks about the strength of the gravitational field at altitude so all these calculations of centrifugal acceleration are irrelevant.

@DavidRicherby: Second paragraph of question: and also possibly a tiny effect from the speed the plane as it travels round the curve of the earth (altitude remaining constant) but does anybody actually know how much less?

The speed of the air is roughly the speed of earths surface (plus wind). Absolute speed of an aircraft is therefore (vectorial) sum of air and (aircraft w.r.t. air).  I've completely rewritten my answer and hope it becomes more clear. If you do not agree, please tell my what the angular speed of my balloon and aircraft is. (by the way: I'm experimental physicist, I'm trying to bring equations and reality together)

Ah, and as I said, traveling west makes the aircraft lighter!

Why are you calculating the angular velocity in such a complicated way? Angular velocity is velocity divided by radius.

Angular velocities can be added just like the linear ones.

just to clear things up (hopefully): for practical calculations, all rotational velocity should be measured relative to the background of stars.  in this case, it is reasonable to measure relative to the surface of the earth since we are seeking a delta, rather than absolute value for centrifugal force.  more detailed discusion here: https://en.wikipedia.org/wiki/Absolute_rotation

Do these curves assume the earth is a perfect sphere? The horizontal black lines seems to indicate so -- but actually the polar and equatorial radius of the earth differ by more than the cruising altitude of the aircraft.

@HenningMakholm yes, I assumed a sphere, because it's easier to calculate. The gravitational field of an ellipsis is more complex.  I can check this later.

@sweber: One could probably get away with approximating the field as that of a spherically symmetric point mass, but the fact that you're closer to that point mass the higher latitude you're at would certainly show up on those plots. (This is not just pure pedantry; I think that taking this into account would show that the range of effective gravity strengths you can experience standing at sea level intersects the range of effective gravity you can experience sitting in an airliner at cruise).

(For actual pedantry, though, I think it would be fairer to model the east and west traveling airplanes not as following a small circle around the pole, but following a great circle that is merely momentarily perpendicular to the meridian -- or at least, for easier calculation, as having the uniform circular motion in inertial space that gives the desired momentary ground speed and direction.)

You are correct that the force of gravity is slightly less the further you get from the earth.   Airlines typically cruise around 30,000 - 35,000 feet.  We can use as a proxy measurement the force of gravity on Mt. Everest, which is 29,000 ft.
The force of gravity on Everest is about 0.434% less than the standard 9.8N/kg.  This means that one pound at Sea Level would weigh about 0.995 lbs. at 29,000 ft.   Or, a typical 180 lb. human would weigh 179.1 lbs.
I don't consider the speed of the aircraft going around the earth to be significant.  Any centripetal force would be extremely tiny.

Do you have a source for your 0.434% figure?

I used a post on this link.  Another source, cites 9.7639m/s-2, which is about 0.436%.  So mutiple sources and values are coming in very similar.

Yeah it sounds reasonable, just wondering where it came from.

"I don't consider the speed of the aircraft going around the earth to be significant. Any centripetal force would be extremely tiny." - unless you're travelling at Mach 10, in which case you're almost orbiting. In fact, can you do the math to see if the effect due to centrifugal forces could be larger than the effect due to altitude? My first guess says they'll be the same order of magnitude.

The question was specifically about "airliners at cruise altitude", which are always sub-mach, and typically around 0.7 - 0.8 @ FL350.  I am not aware of any (unclassified) aircraft that can do M990 within Class A space; They likely do not exist.

I don't consider the speed of the aircraft going around the earth to be significant. Any centripetal force would be extremely tiny. Nope, this effect can be significantly larger!

@sweber: Prove it.

@abelenky: Did it. (see my answer)

The largest delta-G from speed that you described is 0.008145.
The delta-g I estimated from altitude is about 0.04275, or about 5x larger.
The delta-g from altitude is less than 1 lbs per human, and the delta-g is from speed is 1/5 of that, or less than 3 ounces.
I don't see how the speed-effect is significantly larger than the altitude effect.

@abelenky 0.4% looks like a delta-g of 0.004 to me, not 0.04. What'm I missing?

Standard G is 9.80665, minus G @ Everest of 9.7639, for a delta-G of 0.04275 m/s2.     (And that is 0.434% of 9.8m/s2)

@abelenky - The first source you quoted was Yahoo Answers, which is not exactly a reputable source. Your second source (New Scientist) quotes: "Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2". Mount Nevado Huascarán has an elevation of 22,205, well below that of Mt Everest (or typical airliner cruising altitude).

"I don't consider the speed of the aircraft going around the earth to be significant. Any centripetal force would be extremely tiny." And irrelevant: the question is about the strength of the gravitational field, not about any other forces felt by the plane.

@DavidRicherby: The question read, "and also possibly a tiny effect from the speed the plane as it travels round the curve of the earth"

@abelenky In Newtonian gravitation, which you're considering, the speed of the plane doesn't affect the gravitational force at all.

lbs describes  mass not weight, so a 180 lbs man would still have a mass of 180 lbs on the Moon, even if the weight will be 6 time smaller.

@DavidRicherby: but the questioner says, "I'm not talking about a special reduced gravity flight with a parabolic flight path". To me this makes it clear that they are asking about perceived acceleration within a frame of reference fixed to the plane, not about the gravitational force on the plane observed in a frame of reference fixed to the earth. Because the latter is not "reduced" in a parabolic flight. Whether they should have called this "force of gravity" or not is a quibble you should raise on the question, not on answers.

Here's the Fermi Estimate version, in case the math in Aeroalias's answer is difficult to follow:
It's been said that the ISS experiences 0.9 G (90% of standard gravity at sea level), which is of course canceled out by their orbital velocity so the astronauts inside feel like they're in 0G.
Airplanes are said to fly a mile high, and the ISS is over 100 miles high--these are not particularly accurate numbers, but they're good enough for order-of-magnitude estimations.
Therefore, without any complicated math, we would expect an airplane to experience 99.9% of standard gravity.  As Aeroalias's answer works out to 99.63%, this is a pretty good estimate.

Pretty sure ISS is in a zero-gravity environment, otherwise they have a very elaborate cable system to make the astronauts float around...

@RonBeyer: If the ISS would not experience gravity, it would not orbit the earth. The centripetal force just cancles it out.

@RonBeyer: Their orbital velocity balances out the 90% gravity, but it's still there.  (See Newton's Cannonball for the earliest formulation of this concept.)

@RonBeyer You can think of the ISS and the astronauts as falling at the same speed.  But there is surely still gravity- it is gravity that causes them to fall.

I understand centripetal force and how steady-state orbits are achieved, the point I was making is you may want to clarify in the answer otherwise it sounds like there is "gravity" in the sense you feel 90% of earths gravity while being an observer on the station, which isn't quite right.

@RonBeyer Well you don't feel earths gravity when you are falling in a lift at ground level either.

@RonBeyer Edited for clarity

They do have a very elaborate cable system. The whole ISS is on a sound stage in Culver City! ;)

And this is why the terms "zero gravity" or "micro-gravity" so often applied to objects orbiting in free-fall are so damned annoying.  90% of normal isn't "micro", much less "zero".

"Airplanes are said to fly a mile high"  - Uh... NO.  That would be about 5,000 feet.  Planes actually cruise around 30,000 - 35,000 ft, or about 6 to 7 miles

Came here to say that, @abelenky, leaving satisfied!

@abelenky: As noted: "these are not particularly accurate numbers, but they're good enough for order-of-magnitude estimations."

@abelenky  At DEN, planes taxi at a mile (and a hundred fifty feet) high.

@MontyHarder At some airports, the cabin altitude descends after takeoff. :)

@sweber: For an object in circular orbit, gravity is the centripetal force. It doesn't cancel itself out! (However, in a frame that moves with the spacecraft there's a centrifugal force that cancels out gravity).

@HenningMakholm: Yes, of course!!!!

I have to say, the feeling of zero gravity on the ISS has everything to do with how fast they're moving and almost nothing to do with their height. At the right speed, you could experience weightlessness at cruising altitude, or even at sea level (if there weren't air and continents and such to get in your way). So the fact that this estimate comes out with something close to the correct answer is entirely coincidence.

@DavidZ ...and this is why I originally didn't say anything about zero gravity in my answer: because it has nothing to do with what the astronauts feel inside the ISS and everything to do with what the actual force of gravity is that needs to be canceled out by their orbital velocity.  But then someone complained about how of course they don't feel .9 G up there; they're in microgravity, so I edited the answer to account for that.  And now you post this.  *sigh* You just can't win some days...

@MasonWheeler Ah, OK, I see. In that case I take back my previous conclusion, though I still think your answer isn't entirely clear about what it's calculating, or for that matter how it's calculating it. I suppose you took the 10% reduction in gravity on the ISS (compared to Earth's surface), divided that by 100 to account for airplanes being at 1/100th the altitude of the ISS, and then subtracted (10%/100) from 100% to get 99.9%? That wasn't easy to figure out from what you wrote.

Gravity decreases with altitude
See this table providing gravity at different altitudes:

Gravity field at different altitudes (source: The Engineering Toolbox)
At a height of 10 km, gravity is 9.776 against 9.807 at sea level. That's a variation of 0.32%, which I consider as significant from an aircraft design angle, as it allows to reduce the fuel consumption in a larger proportion.
Can we sense the gravity difference at the altitude of 10 km?
Such difference cannot be sensed by a human, a metering instrument is required. Sensing 0.3% difference just requires a scale. If you "weight" a mass of 100 kg, then at a height of 10 km, the scale will just display 99.7 kg.
Note: A gravity value of 9.78 m/s² already exists on Earth, e.g. in Mexico city and in Singapore, due to gravity anomalies.
How fast gravity decreases?
The center of the gravity field is near the center of the Earth. The surface of the Earth is at 6,400 km from the center and the value of gravity is 9.81 m/s² or g.
Each time the distance from the center doubles, the gravity value is divided by 4: At 12,800 km, the value is 1/4 g. This progression is said to be an inverse square law, which looks like this:

Curve of an inverse square law (source)
Many physical quantities are based on this same law (light intensity, sound intensity, radio signal intensity). As you can see after 3 or 4 Earth radius, the variation has slowed down a lot, but it continue to decrease and will never reach zero. it means any object in the universe has an impact on all other objects! (but a small one).
The gravity value decreases when climbing, but also decreases when going underground. Near the center of the Earth the gravity is null (at least that's what we believe, we are not going to be able to check until a very long time, it's easier to explore space than the depths of our planet). This is the complete picture of the gravity:

Gravity field according to the preliminary reference Earth model
Gravity is a puzzling force not yet understood. We know local effects of the gravity, but we ignore the reasons of such effects.

Measuring gravity from the center of the Earth is a little misleading, as all mass has gravity - if you were in the center you would feel a roughly equivalent pull in all directions, not just "downward".  (What "down" even means in the center is up for debate.)  If you dug a really deep hole, you'd thus feel the affect of gravity from the ground below you, but also from the sides and all the mass that's above you.  Thus, your overall downward pull would actually decrease the deeper you went.  A bit pedantic, and not really relevant to aircraft (hopefully), but worth mentioning.

The maximum gravity value is near the center of the Earth (or can be seen as maximum near the center) are you sure that this is the case? Resulting force should zero out as you approach to the center, right?

@Sebi, Yeah, that's what my comment above was getting at.  It would be true if the Earth were treated as a point-source of gravity, but this is obviously a spherical-cows-in-a-vacuum type of oversimplification.  The chart added afterwards clears this up a bit.

@DarrelHoffman: On the other hand isn't our current concept of gravity also oversimplified? Did we see the graviton recently?

Most of questions deal with the gravity reduction due increased distance to the Earth but there is also centrifugal force caused by movement. This force allows spaceships to circle the Earth with no power applied and causes weightlessness inside - just increased distance would not be enough for these effects. An airplane is also flying around the Earth, same as a satellite does, just much slower.
This effect described here and may reduce the perceived weight (the aircraft feels lighter and everything inside is lighter) but can also increase it (depending on how the flight direction is related to the rotation of the Earth). The effect is somewhat about 0.3 % of mass at the speeds close to the speed of sound so comparable to the effect from the increased altitude.

Most answers here are from a perspective of "at a height of X, you weigh Y". Lets turn this around.
At the altitude of the ISS, gravity is about 10% less. I doubt you'd notice 10% reduction in weight (double-blind, if such a thing were possible), without using measuring equipment. That's 400km up (250 miles), and well outside our atmosphere.
The only reason people are "weightless" on the ISS, is that they are in orbit.
Things in orbit are always weightless. The scales you are standing on is falling at the same speed you are. This can happen at any altitude - if you kick a ball, it is in orbit - an orbit that intersects with the surface of the earth, so it can't complete the orbit.

The question is about the effects on an airliner, not on the ISS.

I understand that. My point is that, even at the altitude of the ISS, you'd be hard pressed to notice, so at a plane altitude, which is much less, you definitely wouldn't.

Well, you basically are not answering the question though.

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Was the transatlantic crossing for Concorde too short to reach optimal cruising altitude?
When listening to Gander/Shannon ATC on shortwave, you could hear Concorde communicate its planned flight levels at longitudes from 20 West till 50 West. What I remember is that it would continue climbing to, say, flight level 570 at 30 West and then descent. So, it seems that it never reached cruising altitude, it would climb until halfway on the Atlantic and then start to descent again.

The simple answer is that the Concorde had no single assigned altitude, it was allowed to climb freely above ~FL450; this is discussed in depth in episode 166 – Flying the Concorde (worth the listen as it answers just about every Concorde question!). As @pilothead alludes to in their answer it climbed as it burned fuel but the aircraft never actually initiated a climb, it simply drifted up as it burned fuel and became lighter.
Also discussed in the episode is the complex approach and departure procedure. Due to the fuel burn schedule the Concorde did not really have the ability to hold for more than a single lap in a holding pattern or do a stepped climb with some route adjustments like many airliners. They had a special departure procedure that was more or less runway to cruise with no interruptions and a similar descent option. So the flight was effectively a climb to cruise block then a glide down right to landing. Depending on the wind and conditions of any given day as well as the load on board, the cruise altitude could vary greatly.

The Concorde cruise climb would actually be the most efficient cruise procedure for all aircraft, anyway, but no other aircraft are allowed this due to traffic density. Normal aircraft approximate this by step climbing a couple of thousand feet every few hours. Concorde was very much alone at her altitude, hence a gradual climb didn’t risk any loss of separation with other traffic.

Link for what @CptReynolds said: Step Climb. As the aircraft changes weight the efficient altitude changes. Apparently conventional aircraft used to cruise climb, but since the skies are pretty busy now isn't no longer an option. Interesting read.

Concorde had a 10,000fpm climb and a max altitude of 60,000ft, so time to climb was not a problem. It had an optimum cruise altitude that varied with weight, so as it burned fuel it climbed higher to stay on the optimum.
There were no other aircraft operating at those altitudes, so it would get clearances to climb 15,000ft at a time and would cruise climb throughout the trip until descent to destination was required.

Is there a link to graph/formula of optimum cruise altitude vs weight?

@smci It is more complex than just weight. I found a repository of the flight manuals but it is thousands of pages. If you are interested http://www.avialogs.com/index.php/en/aircraft/europe-and-consortiums/concorde.html

can you just tell us the first-order approximation to the relationship between optimum cruise altitude vs weight?

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Why are airlines so concerned with checked baggage weight?
Why is it that commercial airlines are so concerned with checked baggage weight, such that they charge extra just for the privilege of checking in a suitcase with a limited weight (e.g. 15kg), and then if you should go even a couple of kg over the limit, they charge quite a bit per kg?
My first guess is they see it as a way of making extra money, but is this the only reason?
The reason I'm so puzzled about it is that they don't seem the least bit bothered about the weight of carry-on luggage, not to mention the passenger's weight (e.g. I weigh about 70kg more than my daughter, yet our ticket prices are the same). And then the weight of the aircraft itself - e.g. an Airbus A320 weighs 42.6t, with a max take-off weight of 78t according to Wikipedia. So why are they so concerned about an extra 3-4 kg in the checked-in luggage?

Relevant: https://travel.stackexchange.com/q/52669/12011 (The answers to that question answer this one.)

It is primarily a safety issue. Profit is secondary. Extra 3-4 kg if abused by all passengers on board is around 1/2 a ton.

Airlines do care about your carry-on weight and even your weight. Many have carry-on weight limits, which low-cost carriers usually enforce, weighing the carry-on and tagging it as cabin luggage. Full-service airlines rarely bother.
One reason they don't measure your weight is that most people wouldn't feel comfortable getting weighed in front of strangers, then charged for that. Only Air Samoa currently weighs each passenger and tickets them by weight. In the West, weight is a contentious issue, so an assumed average is used for passengers. Such assumptions have crashed small planes, and in special circumstances like sports charters, actual weight is used. Finnair weighed volunteering passengers to get a better sample.
Weight costs money to carry, and not just in fuel burn. Luggage shares its volume with cargo, which airliners from the A320 up are designed to carry in the hold. Lately, cargo tends to get become more profitable than passenger service. Electronics and other high-value items cost more in depreciation and time value to ship by sea and land than by air, creating massive demand for air freight.
The A320 is 42.6 tons empty, but that's not the number that matters. The number that matters is its 78 tons MTOW, and how short of that it comes after adding pax, luggage and fuel to the 42-ton OEW. That spare weight is money in standby cargo.
Giving potential 175 passengers an extra 20 kg each (long-haul FSA vs average LCC allowance) could, if everyone used it all, add up to 3.5 tons total. Freight rates vary widely, but on a 3,000 mile flight that's worth \$7k-\$10k in freight revenue, or \$40-\$60/pax. Admittedly that's on the upper end of the range, but you can see there's money to be made. For an A320 that only costs ~$7k/hr to fly, it's a lot.
Less ethically, moving weight and volume from luggage to carry-on is profitable for a mixed passenger/cargo airline. Checked-in luggage has to be weighed in all cases, to ensure each piece is within safe limits for manual handling. This known weight then eats into the takeoff weight calculation, leaving less for cargo.
As airliners are designed to take off with one failed engine and carry reserve fuel for failures and diversions, this extra power works as weight safety margin when nothing fails. In practice weight margins rarely get pushed - most flights are volume-limited, with enough weight margin to take off at reduced power. Still, less luggage means more cargo means money.

For pax and carry on, airlines are allowed to assume a standard adult passenger weight of 190lbs in summer and 195 in winter. Carry on bags are assumed to be 16 lbs. (this is in North America).  The numbers are based on assessments of typical weight distributions within a random group of people and carry on luggage.  The 190lb weight used to be 180lbs, but this was increased due to the observed chubbyness increase in the population in recent years.

The average pax weight being too low and lots of heavy baggage contributed to the crash of Air Midwest flight 5481.

"That's why there are carry-on weight limits." Many (most?) airlines don't have carry-on weight limits. Instead, they have a size limit to make sure it will fit in the provided space. It's mostly just LCCs that have carry-on weight limits so they can find an excuse to charge you more. Some mainline carriers do have weight limits for carry-on, but they're typically generous and often not enforced.

@reirab many airlines have both but the weight limit is hardly ever enforced. E.g. KLM has a max weight of 12kg for carry on (https://www.klm.com/travel/nl_en/prepare_for_travel/baggage/baggage_allowance/index.htm) and if at the checkin desk you've clear trouble lifting your carry on bag they will want to weigh it (I've had it happen, with severe back problems...).

@jwenting Yeah, it seems to be more common in Europe and parts of Asia. But, like you said, even then it's often not enforced. In North America, it's uncommon for there to be a weight limit except on ULCCs, enforced or otherwise.

Also note that while passenger won't simply weigh more if allowed to, many will take bigger luggage, so some limit has to be placed on it.

I've been thinking for a while now that you and all your luggage/coats/duty-free etc should stand on a scale and that's the weight each person has a limit for.  Combined weight of everything they're asking to be put on the plane - body included.

@GrimmTheOpiner Seat price plus a weight based fuel charge would indeed be the fairest price structure. But you'd have to refit check-in with flush scales for this. Also, carry-on and especially duty-free weight are difficult to police - weighing and billing passengers at the gate would slow down boarding, cost more gate time, delay flights, and increase turnaround time. Delays at the gate are to be absolutely avoided.

@GrimmTheOpiner Finnair is already weighing passengers. Samoa Air began charging by weight in 2013.

More recently, in 2014 a Qantas jet had trouble taking off because it had 87 children on board that were incorrectly assumed to be adults, throwing off the center of gravity from what was expected.

@JanHudec: Hey, don't be rude to shapeshifting passengers!  :-P

There's a few people using the word "pax", but I never saw what that means?

@JPhi1618 "x" is a sign of contraction, so pax=passengers. https://aviation.stackexchange.com/questions/25162/what-is-the-origin-of-the-term-pax

@GrimmTheOpiner While I was in the US military, some military flights included us grabbing all our junk and standing on a scale. It did slow down things, but not by much.

If an airplane is too heavy when it tries to take off, it might not be able to take off safely (it might still lift off, but be unable to maintain flight in the event of an engine failure).
Additionally, the weights of the things that can render an airplane overweight (mainly passengers and their baggage) are not necessarily centered within the airplane's safe center-of-gravity (CG) range; thus, overloading a plane with passengers and baggage could potentially move the CG beyond its forward or aft limits, resulting in loss of control upon liftoff or shortly afterwards.[1]
The absolute upper limit on how heavy an airplane can be and still take off safely is known, conveniently enough, as its maximum takeoff weight (MTOW); this is the maximum weight at which an airplane is capable of lifting off and remaining airborne, even if it loses one engine at the worst possible moment during the takeoff. Note that the calculation of a particular airplane's MTOW assumes that it will be taking off from a long, dry, sea-level runway with a high local atmospheric pressure, engines at full power, a stiff headwind, and flaps and slats fully deployed; when taking off from a shorter runway, or from one covered in slush, or from an airport at high altitude, or with a low-pressure system over the airport, or with a tailwind, or with noise restrictions that require the use of less-than-maximum power when taking off, what matters is instead the maximum permissible takeoff weight, which is less than the MTOW (how much less depends on all of the above factors, along with any others that would make it harder to take off).
As a result, airlines try very hard to keep their planes from taking off overweight. There are eight major contributors to an airplane's weight:

The dry weight of the airplane itself and of those of its resources that aren't consumed during normal flight (such as hydraulic fluid).
The weight of the fuel and oil carried by the airplane.
The weight of the airplane's crew.
The weight of the airplane's passengers.
The weight of the passengers' checked baggage.
The weight of the passengers' carryon baggage.
The weight of any freight being carried in the cargo hold.
The weight of other miscellaneous consumables (such as the peanuts and soda, henceforth known as "passenger chow").

Here's a diagram showing most of these contributors:

(Image by mins from this post.)
Of these, the weight of the airplane itself is fixed and unchangeable from flight to flight (unless maintenance and/or repair work has been done on the plane in the interval between flights, which could slightly alter this weight). And the amount of fuel and oil carried is determined by the amount of same needed in order to get the airplane from its source to its destination, fly around in a holding pattern for an hour or so, and then fly to the diversion airport listed in the flight plan, plus a generous margin of safety to account for such things as headwinds, fuel leaks, engine failures, or all of the above simultaneously. This leaves six weights that can be minimised:

There are only a few crewmembers onboard the airplane, and there is far less variation in physical parameters among crew than there is among passengers (the geometry of the flightdeck and the controls therein imposes hard upper and lower limits on height, and height is fairly strongly correlated with weight); therefore, although the weight of the crew is known exactly and can be minimised, there is little weight that can be saved by doing so.
The weight of the passengers can be estimated, but, as Therac said, can't generally be measured exactly, for fear of lawsuits by the large of frame.  The total passenger weight can be minimised, if need be, by removing passengers from the airplane, but the airlines avoid doing this if at all possible, because taking passengers off a flight costs them money - and they can't try to offset this by charging heavy passengers more than light ones, again for fear of lawsuits.
Checked baggage can be weighed exactly, and restricting its weight doesn't lose the airline any money (they generally charge passengers based on their number of checked bags, not - up to a certain point, at least - the weight of their checked baggage).
Carryon baggage - see 5. However, as the weight of the average passenger's checked baggage is much more than the weight of their carryon baggage (16.7 kg/passenger versus 6.1 kg/passenger, respectively, according to this study), there is more weight to be saved by minimising the weight of the passengers' checked baggage than by minimising the weight of their carryon baggage (and, as Therac pointed out, the airlines do care about the weight of your carryon baggage - this is the main reason why they limit the number and size of your carryons).
Freight (if any) can be removed to lighten the airplane, if necessary (and its weight can be measured exactly), but, as with removing passengers, the airlines don't like to do this except as a last resort, as it costs them money (since now they don't get paid for transporting the freight that had to be offloaded).
The amount of passenger chow carried is either constant from flight to flight, or else depends on the number of passengers onboard; either way, it can't be independently decreased (at least, not without pissing off the passengers), and the possible weight savings would be minuscule anyway, due to their very small weight.

As you can see, most of the contributors to an airplane's weight either can't be lightened at all (airplane, petroleum byproducts, passenger chow), would incur significant cost penalties in lightening (passengers, freight), and/or wouldn't free up enough weight from being lightened (crew, passenger chow); this leaves baggage (both checked and carryon), and, since the average passenger has more checked baggage than carryon baggage, checked baggage represents a greater opportunity for weight reduction than carryon baggage does.
[1] As happened to, as previously stated in this comment by Michael Hampton, Air Midwest Flight 5481.

"since the average passenger has more checked baggage than carryon baggage"  - at least in Europe, I suspect the median passenger doesn't have any checked baggage - it's all carry-on.  Certainly I have made many more flights without than with.

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text:
What is the heaviest Single Engine Prop Plane that has been flown?
What is the heaviest single engine propeller-driven fixed-wing aircraft that's been produced & actually flown?

Experimental and one-of-a-kind aircraft count, as long as it has actually flown.
The maximum takeoff weight, as certified or designed, is the weight that counts (i.e. not merely empty weight, but with payload + fuel at the start of the flight).
Turboprops count, biplanes count, any timeframe (well, pre-1900 may be hard) and any country, civil or military.

It's worth noting there are single propeller aircraft driven by multiple engines. Does that count? And there are dual contra-rotating propellers driven by dual engines. Does that count?

@mins I'd be good with those - the characteristic I'm most interested in is having a single place on the aircraft with prop blades (i.e. on the nose, not on both wings or front+back), so the number of engines spinning the shaft or the presence of the contra-rotating prop aren't my main concerns.

The Douglas XB-42 Mixmaster had a maximum take of weight of 16,194kg (35,702lbs). (Note - the later A variant had a couple of underwing jets, which took the mtow above 20,000kg.) It was powered by two Allison V-1710 engines each independently connected to one of the contra-rotating propellers. In an emergency, the entire rear assemblage could be explosively jettisoned.

There is the Douglas XTB2D Skypirate with a maximum take-off weight of 15,767kg (34,760 lbs).
It had a single Pratt & Whitney XR-4360-8 supercharged 3000hp radial piston engine driving a pair of 14.3-foot diameter, contra-rotaing four-bladed propellers.

The first U.S. Navy Douglas XTB2D-1 Skypirate prototype (source)
Both these planes have contra-rotating propellers. For the largest plane with a single propeller, it may be the Linke-Hofmann R.II of 1917, with a gross weight of 12,000kg (26,455lbs) . It is said to have had the largest propeller ever flown, its four engines turning a single 6.90m (22 ft 7.5 in) diameter two-blade propeller at just 545rpm.
 (source) looks normal until you notice the people
And... this one is a bit of a cheat, but also the Boeing JB-17G engine test-bed. Since the B-17G had an empty weight of around 17,000kg, it comfortably beats the other contenders.
In flight its main engines were shut down, and it was flown with a single engine.

The XB-42 hat two Allison 1710 engines mounted in its fuselage.

@PeterKämpf  true, but the op says "the number of engines spinning the shaft or the presence of the contra-rotating prop aren't my main concerns" so I think it counts!

Possible candidates:

XF8B with MTOW 21,681 lbs (9,831 kg).

XF-84H? It's based on an airframe with MTOW of 27,000 lbs (12,247 kg) with a turboprop slapped on the front. Wikipedia claims a MTOW weight of 27,046 lbs (12,268 kg), but it's of questionable providence so I'm hesitant to claim it as fact.

Tu-91, MTOW 31,747 lb (14,400 kg). XTB2D, 34,760 lb (15,767 kg)

Grumman Mauler - once lifted payload of 10648 pounds of ordinance alone - empty weight listed as 15,257 lb (6,920 kg), gross weight listed as 25,737 lb (11,674 kg) -- but some heavier listed in comments above


text:
What are the procedures and responsibilities concering landing clearances on ILS Cat 2 or 3?
On very poor visibility, the pilot approaches under ILS procedure.
Usually they are requested to call the TOWER, when visual with the runway, to receive the landing clearance. However, on ILS Cat 2 or 3, it happens quite close to the runway, then the pilots will not have enough time, to switch on the Tower´s frequency to receive the clearance. Then the Approach Control may say: "The tower clears you to land."
My questions are:

Does the tower CLEAR the landing, or only inform the runway is clear?
Is it the pilot´s decision to land or not?
Where can I read this specific document or phraseology?

You call tower when you are visual with the runway if you are doing a visual approach. I don't think it extends to instrument approaches; instead on instrument approach you switch to tower when you are established, which means you are aligned with the specified navaids (usually ILS, but may be MLS, augmented GPS and for Cat I even VOR+DME or NDB).

@Simon That is not correct. With distress traffic or in cases where a clearance can only be issued by ATC not by AFIS for example, the clearance is relayed. IFR clearances at uncontrolled airfields in Germany are always relayed, e.g. "Langen Radar clears you to ZZZZ via ABCDE... blah" in lieu of "You are cleared to"...

Some airports will use Approach to issue Landing (and even Lineup/Takeoff) clearances when LVO are in use. Sometimes we get handed over to tower "for taxi clearance" on arrival, then on departure we are swiched to App frq before even lining up on the rwy. If you think about it, on LVO the tower sees the runway about as much as the App controler.

@Simon during surveillance assisted approaches, clearance to land is relayed by either approach or final control. You then tune up tower on the roll out.

So I was completely wrong.  I've deleted the comment.

Around here we'll typically get our handoff right at the FAF (they seem to know exactly when you're busiest and hit you with more workload right then :-)  ).  The landing clearance comes afterwards.  Sometimes immediately after communications is established, sometimes not until down near DH (depending on other traffic, etc).

Calling the "airport in sight" is an optional call by the pilots and only affects your approach clearance, not your landing clearance.  Prior to any instrument landing you will receive an approach clearance.  A clearance for an ILS approach is the same whether you are flying to category I, II or III minimums.  If you report "airport in sight" to the approach controller they may issue you a "visual approach" instead.
Once you are established on final approach you will be handed off to the tower and they will issue the landing clearance.  In IMC you will receive a landing clearance well before you can see the runway and there is no reason to call the controller when you see the airport.  They will know you can see the runway when you land on it.
Note that the handoff to tower occurs around the final approach fix or when established on the approach, not when you have visual contact with the airport.

You are typically (I hesitate to say always, as I have no evidence for this) cleared to land, and are not requested to call visual with the runway. This is because, as you state, there is not time to call visual with the runway and request clearance: by the time you see the runway you could be at 370m (Cat II) or 0-300m (Cat III) from the end of the runway and less than 200 feet above it: there is not necessarily enough time to call the tower and receive clearance before you would have to miss the approach.
In fact, for a (theoretical, as I do not believe any exist) cat IIIc approach, you do not necessarily even see the runway before your wheels touch it, as there is no required Runway Visual Range specified. In such a case, it would not be possible to call the runway visual.
If the tower clears you "to land", you have permission to go ahead and put your wheels on the runway. There's no ambiguity in the clearance, if you're cleared to land then you're cleared to land. Of course, that never obliges you to land, you're still the pilot in command.

Because, remember, runway visual range is in metres.

A good point, I'm confusing my decision heights and my Runway Visual Range's (ie mixing up the horizontal and vertical components). The point stands, however, that you are cleared to land rather than cleared until runway visual, due to insufficient time to report and receive clearance. And it's theoretical because you can't taxi, true, but that doesn't say you couldn't land one aircraft if needs be (eg in an emergency) and leave it parked on the runway.

Cat IIIc is theoretical, but I also said that in Cat IIIb you may not see the runway before the wheels touch the ground either, because Cat IIIb has DH=0 and that is absolutely not theoretical.

@JanHudec : What does the Runway Visual Range of 150 ft for  Cat III b mean? I thought that mean't you'd see the runway from at least 150 feet away?

@curious_cat, well, effectively it means you can see 150 ft of the runway ahead of you. At normal 3°, which is 5%, glide slope it 150 ft corresponds to height of mere 7.5 ft and you can't do anything in that height. The point of the 150 ft is that you are able to turn off and taxi!

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text:
What is the visual difference between turbine fans and compressor fans?
How to identify that a given fan is a compressor fan or a turbine fan, just by visual inspection? Also, can visual inspection of a fan help in identifying the compressor/turbine stage i.e. if the fan is from a High Pressure Turbine/Compressor blade or Low Pressure Turbine/Compressor blade?

Assuming we are speaking of axial compressors/turbines
Compressor blades are generally thin and straight, and resemble a tiny rectangular wing with low camber thickness.

Image source

Turbine blades are more curved, often in almost "U" shapes (like the ones seen here). In particularly large and recent engines, where efficiency is critical, turbine blades will often be full of tiny holes:

Image source

To distinguish between high pressure and low pressure stages (compressor or turbine does not matter), the length of the blade and its torsion (i.e. how much the aerodynamic profile turns around the axis of the blade going from the root to the tip) are key: shorter and more twisted blades will be high pressure ones, longer and straighter blades will be low pressure.
Note that two blades of the same length could come one from a high pressure stage and the other from a low pressure one of a different engine: "short" and "long" are relative to the engine size.

On the other hand, radial compressors and radial turbines are more difficult to distinguish.

Thanx. Voted up for the description. I'd like to know if there is a way to identify the stage from visual inspection too?

@VictorJuliet if you can look at the blades what's preventing you from identifying it?

Editted the quesion for better understanding.! I am talking about differentiating between the HPT and LPT stages

@VictorJuliet answer updated.

I'm confused; there is no such thing as a "turbine fan". A "fan" is an apparatus with rotating blades that creates a current of air. That's not what turbines do. A turbine converts fluid (water, air, steam, etc.) into rotational movement or power. They are opposites. Someone had to say it. Pedantic

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text:
What is the working principle of tip jets?
Tip jet refers to the jet nozzles at the tip of some helicopter rotor blades, to spin the rotor. What is their working principle?

Tip jets spin the rotor by blowing out compressed air. It's the same principle as a jet engine or a jet-ski impeller, just that the exhaust is pushed out through the rotors. Some designs even had jet engines on the rotor tips! The primary advantage of the design are that there is no torque created, so no need for anti-torque rotors or thrusters, the tip jets also increase the rotor's moment arm and mass, making autorotation safer in some cases, although the drag from the tip jets often slowed the rotors down in many cases. Some say it is easier to learn to fly and safer to fly a tip jet design as there's no torque.
There are very few systems which use tip rotors, the main disadvantages would seem to be a lack of efficiency and drag from the tip jets. The rotor design is complex as high-pressure air must be carried through them, causing many design and materials challenges. Also, lack of a tail rotor means there's no way to pivot in hover unless unless you put some sort of tail thruster on.


text:
How does the spoiler actuation mechanism work on the Airbus A340?
How does the servo control spoiler actuation mechanism on the Airbus A340 work?

Don't limit yourself to searching on the A340. The A330 has a very similar wing and correspondingly similar systems.

I'm a bit unclear on what you are looking for... the design of the actuation mechanism, spoiler flight computer or communication protocol between the two?

Design of the actuation mechanism

If I dont limit to A340, the answer might be marked as : too broad :D

@VictorJuliet, are you really asking about design of the actuation mechanism? Because you accepted an answer that does not even mention it.

Most aircraft utilize a weight on wheels switch. When the landing strut is compressed when ground contact is made, the weight on wheels switch is activated. The signal is distributed to various systems in the aircraft, including the spoiler system.
The spoiler is activated as soon as the aircraft touches the ground to "scrub off" any lift and to make sure the aircraft doesn't become airborne as the jet engines spool down.


text:
Basic aircraft performance: how can you calculate the required power?
How can we determine the minimum power required for a propeller aircraft?

How can we determine the minimum power required at standard altitude of h=3km

and

What is the maximum airspeed at standard sea level conditions ?
if :

Wingspan = 10m
Wing area =15m Squared
Mass = 1500 kg
Power plant = one-piston engine of 230hp power available at sea level
Parasite drag coefficient= 0.025
Oswald efficiency = 0.8
Air density at 3 km = 0.9093 kg/m3

if you are the same person, you could have edited your previous question

I'm voting to close this question as off-topic because If you are going to get help on blatant homework type questions the best you can really hope for is for someone to point out the error you are making in your reasoning or work. Without seeing your work, we cannot do that. We will not just do your work for you.

anyways thx 4 ur advise sir

Hello Mohammed, welcome at aviation.SE. This looks like a homework exercise. That's not a problem, we can help you to understand the fundamentals behind the questions. Please show what you have done so far, and where you get stuck. We will not just solve your assignments for you to hand them in. In the end, the purpose of studying aerospace engineering is to become a good engineer with a prosperous future; if we  would just solve your homework we would do you a disservice.

Hello dear friend DeltaLima and thank you for your comment actually i am studding to be a pilot and i am not going to be engineer in the future so that is why i do not need to focus in this subject however this is the final take home exam and the last deadline is tomorrow morning and to be honest with you tomorrow i have an Air Law exam and i want to focus on it bcz its more important than this one  that is why i shared it here and if there is someone who can help me i would be thankful for that THANK YOU again :)

@MohammedNouri This is the lectures series you need.

@MohammedNouri See also this formula sheet and these formulas. If you need more reading material (slides and stuff), send me a email, address in in my profile page text.

Hints for your first question:

Power required is the product of drag and true airspeed.

$P_r = D\cdot V$

You can calculate drag as a function of airspeed.

$D=\frac{1}{2}\rho V^2 S C_D$

The drag coefficient is the sum of the parasite drag coefficient and the induced drag coefficient.

$C_D = C_{D,0} + C_{D,i}$

The induced drag coefficient is a function of the lift coefficient, the wing geometry and the Oswald efficiency factor.

$C_{D,i} = \frac{C_L^2}{\pi A e}$

The lift coefficient can be calculated as function of the speed, the density of the air, the weight of the aircraft and the wing area.

$C_L = \frac{2 L}{\rho V^2 S}$ and for horizontal flight $L = W = mg$
Substituting yields:
$P_r = \frac{1}{2}\rho V^3 S C_D = C_{D,0}\frac{1}{2}\rho V^3 S + \frac{2 W^2}{\pi A e \rho V S}$
Now you need to find the minimum of that function, which occurs when it's derivative equals zero.
All the constants you need are given in the table, apart from the density at 3 km altitude, but I believe you have been given that as well.
The same power formula derived above will also help to solve your second question.

authors: Виталий Шарагин


text:
How often are aircraft flown by two FO's?
On Air France flight 447 the captain was on rest break and the a/c was being flown by two first officers. Most other instances I know of on long flights there is a relief crew of both a captain and an FO. How common is it to have two FO's and no captain in the cockpit?

Related: How many pilots do long flights need?. "on long flights there is a relief crew of both a captain and an FO". Not necessarily, see the question.

What are FO's??

@Ethan FO=First Officer

I don't think its very common especially on the 747-8 which carries 2 pilots instead of 4.

Keep in mind that, despite their titles, the first officers are fully qualified pilots and are just as capable (and certified) of flying the aircraft as the captain.

@Ethan Beg pardon? 747-8s have relief crews just like all other long-distance A/C.

@Lightsider Apparently not in the case of AF 447. They had a fairly simple and temporary malfunction and they crashed the plane. In the very detailed analysis I read this played a role. The two FO's went back and forth over control because there was no clear command structure.

@Lightsider That's more true in some parts of the world than in others. The U.S. is much more strict about who is allowed to fly an airliner than most other countries, especially in regards to the SIC (i.e. FO.)

I'd say this is quite common on flights that require at least 3 pilots due to length.  The third pilot is almost assuredly going to be qualified as a first officer and this is the common staffing strategy with the US carriers that I am familiar with.
Three pilots (by US flag rules) are going to be required for 8-12 hour flights and you'll find that US carriers often will staff a third pilot even on the shorter US east coast <-> western Europe flights.  But what must the three pilot composition be?  Lets examine 14 CFR 121.543 the requirement to relieve a PIC (excepted):

§121.543   Flight crewmembers at controls.
  (a) Except as provided in paragraph (b) of this section, each required flight crewmember on flight deck duty must remain at the assigned duty station with seat belt fastened while the aircraft is taking off or landing, and while it is en route.
(b) A required flight crewmember may leave the assigned duty station—
(3) If the crewmember is taking a rest period, and relief is provided—
(i) In the case of the assigned pilot in command during the en route cruise portion of the flight, by a pilot who holds an airline transport pilot certificate not subject to the limitations in §61.167 of this chapter and an appropriate type rating, is currently qualified as pilot in command or second in command, and is qualified as pilot in command of that aircraft during the en route cruise portion of the flight. A second in command qualified to act as a pilot in command en route need not have completed the following pilot in command requirements: The 6-month recurrent flight training required by §121.433(c)(1)(iii); the operating experience required by §121.434; the takeoffs and landings required by §121.439; the line check required by §121.440; and the 6-month proficiency check or simulator training required by §121.441(a)(1);

The limitations of 61.167 mentioned above mean those under the age of 23 are not eligible to relieve the PIC. What does it mean to be "qualified as pilot in command of that aircraft during the en route cruise portion of the flight?"  The closest I can find to answer that is in 121.436 which says to be act as a PIC you must have an ATP and be at least 23 years old, have a type rating and have at least 1000 hours combined from part 121 SIC, part 91.1053(a)(2)(i) PIC, part 135.243(a)(1) PIC time.
What this means practically, is that a second in command who

holds an ATP,
is at least 23 years old,
is type rated in the airplane,
is qualified as a second in command, and
has at least 1000 hours SIC time with any airline

may relieve a PIC during en-route flight.
This is further reinforced by the remainder of the quoted portion of 121.543 that says an SIC qualified as PIC need not have any of the required checks a PIC must have (line check, 6 month sim check, etc).
At a major airline it is likely that SIC have ATPs and are type rated during initial training and most major new-hires will be older than 23.  It is also likely they have prior airline experience to give them 1000 hours SIC.  Even if they don't have that experience, they'll have it in a 18 months or less after being hired.
This means that pretty much any FO at the airline and surely any FO who has been there for a year or two is going to be qualified to relieve a PIC.  Because an FO can relieve the PIC and can also relieve an SIC, the third pilot need only be an FO.  This means our crew of three will be one captain and two first officers.
Given the economics involved and the behavior of US airlines when it comes to costs, they will staff an FO wherever they can versus staffing a captain.
To bring this full circle, back to our 8-12 hour flights to/from the US by US carriers that require 3 pilots they will almost always be flown by two first officers during the en-route portion of the flight.

I'm not familiar with non-US regs, so this might not apply.  Given that air carrier rules by the FAA are compatible with ICAO though, I'd guess that this is the common case in other jurisdictions as well.

That makes economic sense. Since an FO has the same technical qualifications as a captain. It just turned out badly on AF447.

The non-US regs have lower requirements for SIC, but the requirement that to relief PIC the qualifications for PIC are needed should be the same.

It happens often, and the one with more seniority sits on the left seat and is Senior First Officer.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

I thought it answered it pretty well.

Perhaps you can expand the answer? At its current state, it is quite short and would be more suitable if it were a comment.


text:
Do crosswinds affect which pilot lands the plane?
I know the crew usually decides ahead of time which pilot is to be PF for each phase of the flight. If, when they arrive at their destination they discover a considerable crosswind component, does that ever change that decision? It seems that the upwind seat would have a considerable visual advantage. (or maybe that isn't true?) Also, do airlines have guidelines on things of that sort requiring that the PIC fly the landing?

The pilot with more experience would usually land in a heavy crosswind situation.

Never ask a question with "ever" in it - because the answer will very rarely be no!
For example - if one pilot is training, then while they may have planned to do the landing reality might dictate that they have to hand over the controls to the instructor. I'm sure the same will apply to type ratings or differences training - nobody is going to let you have a crack at a severe crosswind landing during your first front-seat flight in an A380.
Additionally, flying schools will often have strict crosswind limits and some may not grant you full "As Pilot Operating Handbook" privileges until you have [x] number of hours AFTER your test.
In other words - you may have two qualified pilots, but only one of which is allowed to land the plane!
But I'm being deliberately pedantic - I suspect you're focussing more on commercial, multi-crew operations rather than two pilots who happen to be in one aircraft and I'm not qualified to answer that bit!
I'm not aware of any requirement to take seat position into account, however.

Yeah, i know the pirfalls of absolutes, especially in aviation. Sometimes hard to avoid words like "ever, "never, " and "always." But that said, you gave a valid answer. In the cases you mention are certainly factors that would affect the decision.

There is almost always some crosswind component on every landing if there is any wind at all.  Therefore the pilot who is making the landing, either left or right applies the necessary corrections to make a successful landing.
Crosswind landings is taught to the beginning pilot early in his initial training to cope with the problem.  Each class of aircraft has different approach and touchdown methods to cope with the cross wind problem.

Hello there! This does not seem to answer the question: Would the crew ever change their decision on who's landing based on a stronger crosswind component? Is the sight of one of the pilots significantly limited due to crosswind?

There is one Pilot in Command, it he who decides who is controlling the airplane, therefore the decision as to who is best qualified to make the landing for all existing conditions is up to him.

Please, have you even read the question and my first comment?

A crosswind landing appears the same from either seat.  The pilot has to control the airplane to make land in the direction of the runway regardless of the wind or he will go off the runway in to ???   A pilot does not look only down the runway,

Draw a runway on a piece of paper and place it on a table.  Now notice how it changes shape as you walk toward it without stooping over.  Next approach the table with your head almost a table level and notice the appearance changes.  Step pack a few paces and line up with the runway, imagine yourself on a glide slope as you step toward the runway.  It you stepping to  the right slightly, staying on the imagined glide sloop,  you are seeing what a left crosswind approach looks like.

This answer is saying that every pilot is trained to land with a crosswind, almost every landing has some crosswind, and aircraft are designed to land with a crosswind, so there is no issue with the pilot being on the wrong side of the aircraft. It could be more explicit but I think it does answer the question.

I wasn't sure if there was a significant difference in visibility. Just seemed like if you were crabbed the one on the upwind side would have a better view. But I've never experienced a significant crosswind landing from the cockpit. [If i did i probably would've wet myself! To an untrained eye they look a lot hairier than they apparently are.)   I'm sure there are crosswind limits for aircraft and i would suppose airlines probably impose their own limits. Although both pilots are fully qualified and capable, I wonder if there are circumstances where airlines mandate that the captain be PF

One side would have a slightly better view than the other during a crab, but not by enough to trump skills. The pilot with more experience would handle any unusually difficult situations, including high crosswind landings, regardless of who was on the "better" side.


text:
Is the Bendix/King KMD-550 airplane icon always oriented to the north?
I have a 2001 Cessna 182T with King avionics. The KMD-550 moving map display correctly shows "track up" but the little airplane icon is always pointed to the north.
How can I set this so that it points track up as well, or is there some kind of fault with the unit?

Does the aircraft icon rotate if the map is set to "North Up"?

Can you please add a picture? This PDF (p.18) states that Aircraft Symbol - Indicates present aircraft position and heading (if available) or track. Do you require some additional circuitry or subscription to show actual heading?

This is really a question you should be asking Honeywell/Bendix-King (or your dealer), but do you mean "points north" or "points up"? As I understand it the latter is expected behavior with the KMD-550 in "track up" mode: The airplane's nose will point along the current GPS track (or heading, if you have a heading input), which is "up" on your display in "track up" mode. If the nose is pointing to "north" rather than along your track you may have found a bug, but only Honeywell could say for sure.

does it say "no hdg or trk" at the top of the screen?  or perhaps "[mag/tru] ### trk"?  if the first, your unit is not getting heading data.  if the second, your unit is functioning normally.

@voretaq7 - I think you've got the correct answer, which is where I was heading (heading, get it?). If you want to put that in an answer form...

I'll take a picture when I fly this weekend. The airplane icon is always pointed north - not up. The map is arranged track up. I didn't see any warning of "no hug or trk" but I will look. I've got my avionics person looking at it right now and he thinks it might be a problem with the heading source because the traffic system is giving heading source failure errors as well. I'll update when I know more.

@voretaq7 Wanna write an answer so this disappears from the Unanswered list? :)

Here is what we've done: removed WX-500 to see if that was confusing things (no), used another KMD-540 unit to see if that was bad (no), checked the X-Y-Z lines 3 times (ok - apparently). The only thing left to do is to replace the KTA-810 with a known good unit and that hasn't been possible yet. I'm trying, believe me! One note though - a workaround is that if you're on the traffic page, it uses the antenna's orientation so the data is correct. Not great and not useful with the map, but better than nothing. I'm still working with my avionics tech to solve the problem. Thanks and stay tuned.

According to the Source:

The bottom is for the icon in track up mode. For the North up mode, It
  Depends on the Heading. The fix icon should be placed on one of the 8
  positions (up left, up right, up center, bottom left, bottom right,
  bottom center). For example, if the Heading is 45°, the fix icon is
  placed on the bottom left part of the screen.

I noticed that I had posted this a while ago. Short answer: equipment failure. Each component in the system has its own internal heading source:
- Stormscope
- HSI
- Traffic advisory system
If one of these doesn't work - the little airplane points the wrong way. There is no visible failure message. In fact, I replaced the KMD-550 with a Garmin GTN 750 and the problem stuck around. I had to send the traffic advisory system away for a very expensive repair... wish I had known before I bought the plane.


text:
How is the induced drag calculated for a wing with elliptical planform?
How is the induced drag calculated for a wing with elliptical planform ? Is this wing shape the most efficient ?

@mins The most induced drag-efficient according to Lifting Line Theory, but other shapes can be more efficient overall when wing weight and other things are considered

Induced drag is caused by the downward deflection of the air streaming around the wing. The resulting aerodynamic force is tilted backwards by half the deflection angle, and the air flows off the wing with an added vertical speed component, producing downwash. Increasing the downwash angle means increasing both lift and the backward tilt, so the induced drag goes up with the square of the lift produced.
If you want to minimize induced drag for a given lift, this quadratic dependence means the optimum is reached when the downwash angle is constant over span.

How is the induced drag calculated for a wing with elliptical planform?

The elliptical, untwisted wing has the same angle of attack and the same lift coefficient over span, and produces the desired constant downwash angle.  To simplify things, let's assume the wing is just acting on the air with the density $\rho$ flowing with the speed $v$ through a circle with a diameter equal to the span $b$ of the wing. If we just look at this stream tube, the mass flow is $$\frac{dm}{dt} = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v$$
Lift $L$ is then the impulse change which is caused by the wing. With the downward air speed $v_z$ imparted by the wing, lift is: $$L = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v\cdot v_z = S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho$$
$S$ is the wing area and $c_L$ the overall lift coefficient. If we now solve for the vertical air speed, we get $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$
with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. Now we can divide the vertical speed by the air speed to calculate the angle by which the air has been deflected by the wing. Let's call it $\alpha_w$: $$\alpha_w = arctan\left(\frac{v_z}{v}\right) = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$
The deflection happens gradually along the wing chord, so the mean local flow angle along the chord is just $\alpha_w / 2$. Lift acts perpendicularly to this local flow, thus is tilted backwards by $\alpha_w / 2$. In coefficients, lift is $c_L$, and the backwards component is $\alpha_w / 2 \cdot c_L$. Let's call this component $c_{Di}$: $$c_{Di} = arctan \left(\frac{c_L}{\pi\cdot AR}\right)\cdot c_L$$
For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the reaction force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR}$$
If the circulation over span has an elliptic distribution, the local change in circulation times the local amount of circulation is constant, and the induced drag $c_{Di}$ is at it's minimum. If this would be different, a higher local $v_z$ causes a quadratic increase in local induced drag, so the whole wing will create it's lift less efficiently.

Is this wing shape the most efficient ?

Only if you ask an aerodynamicist would the answer be yes. An elliptic wing will give you the best ratio of lift to drag, which clearly is one way to express efficiency.
In reality, the wing has to lift itself plus a payload, but only lifting the payload should be considered when formulating efficiency. Therefore, pure lift/drag optimization is too narrow. What should count is the best ratio of lift minus wing weight relative to drag. R. T. Jones wrote a NACA Technical Note back in 1950 in which he looked at this problem analytically. Wing weight goes up when much lift is created near the tips, because this lift will cause a disproportional root bending moment, and the wing spar which has to carry this bending moment is a significant part of the wing structure. Therefore, reducing lift at the tips and adding more lift at the root will create a lighter wing for a modest drag increase, resulting in an overall optimum for an almost triangular lift distribution. When compared to an elliptical wing planform, the total wing span of such an optimized wing is bigger for the same overall drag, but this wing will weigh less.

Spanwise loading comparison for wings of the same fixed lift, from NACA Technical Note 2249.
But this is too easy. Scaling laws must be considered in addition. You know that elephants have much more massive legs relative to their body size than antelopes (or even ants, for an even more drastic comparison), since body mass scales with the cube of linear dimension while structural strength scales only with the square of linear dimension. This means that wing spar weight will be proportionally higher for larger aircraft.
As a consequence, insects have more elliptic wings than albatrosses, and model aircraft have optimum wings which are much more elliptic than the optimum wing of an airliner. The optimum shifts from an elliptic load distribution at very small scales to an almost triangular distribution at large scales.

Why in computing the lift from downwash did you assume that the wing influenced uniformly just a circle of air of diameter equal to the span ? I would think that the downwash effect of the wing would be variable from near to far away.

@EarlQuandt: Yes, in reality the influence of the wing fades the more you move away from it. But this is a simplification which you will also find in propeller theory, and it gives the correct result, because what you neglect in the far field you compensate by assuming a constant influence over the near field.

I am uneasy about calculating induced drag from a model of the wake that is not really there. I thought the wake field of the elliptic wing before roll-up was more like a flat sheet.

@EarlQuandt: Yes, there is a flat vortex sheet in potential flow. But there is no such sheet in reality, only parcels of air. An untwisted elliptic wing will have all parcels of air flowing off the trailing edge moving downwards at the same speed, and also those above and below. The movement will extend up- and downwards more where wing chord is higher. In the center of the elliptic wing a bigger air mass is affected per unit of time than at the tips. This is nicely reflected in my simplification. Your uneasiness comes from the simplifications and misinterpretations of potential flow theory.

Mr Kämpf is trying to refute real world wind tunnel tests with unsupported personal philosophy. What he refers to as "scaling" is Reynold number and the statement that " almost triangular distribution at large scales" is false; look at the Antonov An-225 (worlds largest), C5 Galaxy, or  Spruce Goose. Any similarity to a triangle is simply structural limitations and fuel requirements of a high aspect ratio which is always the most efficient wing shape for subsonic

For a wing with elliptical span loading, the induced drag can be directly calculated from the  lift coefficient. The induced drag coefficient $C_{D_{i}}$ can be calculated as,
$C_{D_{i}} = \frac{C_{L}^{2}}{\pi A}$
where $C_{L}$ is the lift coefficient and $A$ is the aspect ratio.
Elliptical loading produces the minimum induced drag according to the lifting line theory, when only the span and lift are considered. If other considerations come into play (like wing bending moment), the most efficient shape varies.
As for why the best distribution is elliptic, the equations can be readily derived from lifting line theory; basically this is because the downwash is constant along the span. A good way of reasoning why this is so is given in The Minimum Induced Drag of Airfoils by Max Munk, NACA Report No. 121.

If the distribution is the best one, the drag cannot be decreased or increased by transferring one lifting element from its old position to some new position. Now, the share of one element in the drag is composed of two parts. It takes a share in producing a downwash in neighborhood of other lifting elements and, in consequence, a change in their drag. It has itself a drag, being situated in the downwash produced by the other elements.
... In case of the lifting straight line, the two downwashes, each produced by one element in the neighborhood of the other, are equal. For this reason, the two drags of the two elements each produced by the other are equal too, and hence the two parts of the entire drag of the wings due to one element. ... hence, the entire drag due to one element is unchanged when the element is transferred from one situation to a new one of the same downwash, and the distribution is best only if the downwash is constant over the whole wing.

For this reason, when only the span and lift are considered, the elliptical loading gives the minimum induced drag as the downwash is constant over the wing. When the constraints are modified, other distributions and wing shapes become more efficient. For example, from On the Minimum Induced Drag
of Wings by A. H. Bowers:

$\diamond$ Prandtl/Munk (1914)
Elliptical
Constrained only by span and lift
Downwash: $y = c$
$\diamond$ Prandtl/Horten/Jones (1932)
Bell shaped
Constrained by lift and bending moment
Downwash: $y = bx + c$
$\diamond$ Klein/Viswanathan (1975)
Modified bell shape
Constrained by lift, moment and shear (minimum structure)
Downwash: $y = ax + bx + c^{2}$

I don't understand the formula proposed by Klein/Viswanathan, I assume x is spanwise coordinate, so why is there ax, and bx? Also, what is the benefit of squaring a constant?
I suspect there is a mistake, because If we reduce the formula we get the linear equation of Horten.

I checked the source you indicated, and it's indeed an error in the conclusion slides. See slide 16 for the right formula: “Hence the required downwash-distribution is parabolic.
y = ax^2 + bx + c"


text:
Are there regulations about objects/buildings above (across) taxiways?
ICAO doc9157 part2 recommend many standard of taxiway system, including width, separation, slope,etc.
How about height of the object above taxiway, like passenger bridge or rail track? Is there any ICAO doc recommendation or major aviation authorities' rule(like FAA) about the vertical clearance of taxiway?

Just guessing, section "CS ADR-DSN.T.915 Siting of equipment and installations on operational areas", page 115 of the EASA document "Aerodrome Design"may applies also for vertical clearance.

The FAA's airport design guidance has an entire chapter on bridges, but it only talks about bridges that are part of taxiways. There's no information at all about bridges over taxiways, at least in that particular doc.

Bridges over taxiways are not a common airport feature. The FAA does have standards for wingspan clearance on taxiways, but not heights. Airport information will tell pilots if there are wingspan or height limits on certain taxiways, and ATC will know which aircraft are or are not able to take certain routes.
For example, Denver International Airport has a bridge over one taxiway. This provides 42 feet of tail clearance, which leaves 3.5 feet of clearance for the A320 family (except the A318). These are the typical aircraft that use this taxiway. A 737 classic would have 5.5 feet of clearance but the 737 NG and A318 would only have 1 foot of clearance. Since the 42 foot number is only in the center of the taxiway, and accounting for other conditions, 1 foot of clearance does not leave much margin for error. So for regular operations, the A320 and 737 classic are the largest planes that will pass under the bridge.
The bridge at Gatwick Airport provides 22m of clearance, enough for a 747 (19.4m high) but not an A380 (24m high).
This suggests that about 10% margin is enough clearance for taxiway bridges. The FAA requires that taxiways have 20% margin plus 10 feet to objects. While wingtip clearance has many variables that require larger margins, tail height will not change much.

In effect, the FAA's Airplane Design Group specifications impose a lower bound on how high bridges over taxiways must be if a taxiway is to accommodate aircraft of a given Design Group. The Airplane Design Group imposes an upper limit on both wingspan and tail height for airplanes in each group, so an airport designing a taxiway to accommodate a given group would need to make sure the clearance above the taxiway at least exceeded the upper limit of the tail height for that group.
Here are the tail height limits for each group. Each design group goes up to, but does not include the upper limit.

Group I: 20 feet
Group II: 30 feet
Group III: 45 feet
Group IV: 60 feet
Group V: 66 feet
Group VI: 80 feet

Source: Table 1-1 of FAA Advisory Circular 150/5300-13 (PDF), located on page 15 of the PDF.
To get a feel for what these limits allow, here are some examples of aircraft in each design group:

Group I: Most light GA aircraft
Group II: Most regional jets
Group III: Boeing 737 and Airbus A320 families
Group IV: Boeing 757 and 767
Group V: Boeing 747 (up to -400,) 777, and 787, Airbus A330, A340, and A350
Group VI: Boeing 747-8 and Airbus A380

I am not, however, sure by exactly how much the clearance underneath the bridge must exceed the ADG tail height standard. I'd expect there's some guidance for this, but I wasn't able to find it in a quick search of that document. If anyone does find that, feel free to edit it in here.


text:
What are the safety concerns and efficiency benefits of two in-line runways?
Heathrow Hub is an independent proposal to expand capacity at Heathrow Airport by extending both of the existing runways up to a total length of about 7,000 metres and dividing them into four so that they each provide two pair of full length, in-line runways, allowing simultaneous take-offs and landings.

How to ensure the planes landing and take-off in the same runways safety?
How many flight can be handled by such design, and how efficiency it is compared to two runways using segregated parallel operations (one for landing and another for takeoff)?

I get what RedGrittyBrick is saying, but my response to him is "we should speculate on what these impacts are on flight ops". However, that leads us to a question closure as primarily opinion based. On the other hand, since this has never been done before, we have only speculation or (if someone can find it) access to Heathrow Hub's studies on safety/efficiency/capacity/etc. I like this question, but I'm not sure we can answer it within the SE guidelines.

It's not really the same scenario, but there's some similarity to the setup for Oshkosh, where they famously mark multiple landing points on their runways using colored dots, to fit in as many aircraft operations as possible simultaneously. It's not at all the same thing as airline operations, of course.

The Heathrow Hub project hasn't published their detailed safety analysis, although RedGrittyBrick's links do touch on the issue.  I think the bigger question re: safety, though, is "how would you set up the missed approaches to avoid conflicting with the departures, especially if someone goes around waaaay late?"

How to ensure the planes landing and take-off in the same runways safety?

No one knows.
Note this isn't exactly what is being proposed

By dividing it in two with a generous central buffer zone, we create two in-line runways, allowing aircraft to simultaneously land on one and take off on the other.

So that isn't simultaneous landing and taking off on the same runway.
Currently only one LHR runway is used for take-offs at any time. The runway is switched to provide respite from noise to local residents in the take-off path.
Also the surrounding airspace isn't set up in a way that would allow separation for concurrent take-off from two runways. A major reorganisation would be needed of airspace in that region of the UK.
I believe that currently, simultaneous take-off require parallel runways a mile or more apart.

How many flight can be handled by such design

No one knows. Maybe a c 46% increase from 480,000 to 700,000 air transport movements a year. See reference below. Bear in mind that Heathrow is subject to a night-time curfew.

How efficient is it compared to parallel runways?

No one knows.

According to the airports commission in 2014:

The concept of an in-line runway is unconventional, has never been used elsewhare and is not envisaged in international, EU or national regulatory instruments, standards or recommendations for safe design or operation.

They do this sort of in-line runway operation every year at Oshkosh, minus the "buffer zone" (this year's NOTAM is available here). It can be done safely, but i's definitely "unconventional".

And if the landing aircraft overruns the runway and crashes into the aircraft waiting to take off on the next runway in line?


text:
What airport proposal(s) has London seen after WWII?
Instead of Thames Estuary Airport from Maplin in 70s to the latest inner Thames Estuary Airport, is there any proposal airport(s) raised after WWII but cancel/rejected eventually?
I know there is a airport proposal in the north-west London but it is cancelled eventually. Does anyone know about this?
In stead of these two, is there any other airport proposals?

I found the nw airport proposal in Cublington, Buckingham in 60s

There was one more proposal before Thames Estuary Airport. That was the Cublington airport, which was rejected by planner Colin Buchanan,as "an environmental disaster." Later the government rejected the proposal of this airport, and rather went for Thames Estuary.
Other than that, there are no more proposals for new airports, but there are certain plans for expanding existing Heathrow and Gatwick.
The Airports Commission says there is a need for one additional runway in the South East of England by 2030.
It has shortlisted three options:

A third runway at Heathrow
Lengthening an existing runway at Heathrow
A second runway at Gatwick

Read in detail about all of them HERE

There's also the expansion of Southend's commercial activities.

Cublington airport in 60s is new

@Him Thanks  :)..


text:
If I do PPL ground school online, do I have to do all of it?
I am not doing normal ground school (I chose to do it online). I have passed my PSTAR, and will be doing my ROC-A in the upcoming weeks.
With this in mind, must I do the entire online ground school course? I do not know if my instructor (or worse, the FAA) will check if I have finished all of the online work.

Forgive me, but are you sure that your attitude and flying are compatible?

Just a brief reminder to everyone: Voting a question up or down is a comment on how good/bad the question is, not how good/bad the attitude of the questioner themselves is.  This question is actually a very important one, and it's written well.  I'm not sure why anyone is voting it down.

You need at least 40 hours of ground school, according to the Transport Canada regulations (421.26). So you will need some sort of evidence to show that you have done so.
I do hope that you weren't seriously considering doing this. Please read the answers to this question.
You will see a common theme - most accidents are caused by the pilots themselves. Knowing aviation theory inside and out is the best protection you have against joining them in those statistics. It is not a stretch to say that your life depends on the attitude you bring to your flight - starting from the first lesson of ground school.


text:
For an FAA commercial license, can I use EASA/ICAO 2nd class medical?
I have an FAA 3rd class and EASA 2nd class medical. On my EASA 2nd class medical it says "This medical certificate complies with ICAO standards".
To obtain an FAA commercial pilot license, will I need to re-take the 2nd class medical by FAA standards with a different AME or does the FAA accept an "ICAO compliant" medical?

FAA medical certificate requirements are:

Medical certificates are designated as first-class, second-class, or
  third-class.

First-class is designed for the airline transport pilot
Second-class for the commercial pilot
Third-class for the student, recreational and private pilot

Coming to your question, this site states (my highlighting):

... the ICAO naming structure for the various classes of medicals does not
  fit with that of the U.S. Under the ICAO definition, a second class
  medical applies only to holders of private pilot licenses.
In most States, outside of the U.S. and Canada, it is easier to simply
  carry a valid first class medical rather than trying to explain to a
  foreign inspector that a FAA second class medical equals an ICAO class
  1 medical.

For more information, please read this PDF.

This is good information, but it seems to be about using an FAA medical for international flying, not using an ICAO medical to get an FAA license.

Yeah, I forgot about the EASA only having one medical for both commercial and ATP. So the EASA 2nd class is worth nothing, unfortunately. Which i stupid because it was a much more thorough examination. Got my EKG taken which was not needed for the FAA medical at all.

There's no provision in the FARs for using a foreign medical in place of an FAA one. The only place it's mentioned is under 14 CFR 61.75, where you can apply for an FAA foreign-based private license using a foreign license and medical.
But you don't need a second-class medical to get a commercial license, you only need it to exercise commercial privileges (see 14 CFR 61.23(2)(ii)). If you just want to get the commercial license as a personal training/development achievement then you can use your existing third-class medical, but if you plan to use your new privileges then you do need a second-class one.

You can "obtain" an FAA Commercial Pilot Certificate with a Third Class Medical Certificate; the Second Class Medical Certificate is only required to be able to "exercise the privileges" of a Commercial Pilot.


text:
Why was I getting an uncomfortable feeling in an R22 at 3,000 feet?
I was thinking through my flying experiences when I remembered a strange sensation while I was a student.
In a Robinson R22, with about 35 hours experience, I was doing a solo flyout to a nearby airfield for a cup of coffee and a look at the weekend traffic.  To get there, I had to cross a MATZ which included some permanent danger areas up to 2,500 feet.  I'd transited this MATZ several times with a call to the tower but on this particular occasion, I couldn't reach anyone.
Being quite inexperienced, I didn't know if it was just unmanned, I had the wrong frequency or something else so I decided, with no particular place to go, to climb to 3,000 feet and go over the top.  One lesson learned is how long it takes to get to 3,000 feet in a Robbo, especially in a climbing spiral :)
Anyone who has been in an R22 will know that the visibility is superb and as I climbed, the views were spectacular.
As I got to 3,000 feet and leveled off, I had this strange sensation.  It felt as if the aircraft was balanced on top of a tall pole and the cabin had shrunk to really small. I felt as if the aircraft might tip off the top of this pole and plunge.  I also found it difficult to believe the 70kts indicated since the apparent speed over the ground was so low which makes sense given that I'd only ever been that high as a passenger.
I really didn't feel comfortable although I wasn't scared, didn't have any anxiety and felt fully in control of myself but I did abandon the plan, got back down to the 1,000 feet I was more used to and completed a couple of circuits at my home field before landing.
What did I experience?  Has anyone else had anything like this?  I guess that it might be a symptom of vertigo but I have never had problems before or, perhaps something weird in my ear canals following the long spiral climb followed by a level out?

Unusual visual cues probably. I had a similar experience years ago in an ASW-20 in almost 2000m. I felt fine until I started to cross the Vättern, a big inland lake in Sweden. I felt very uneasy and was scared until I had land again under myself. I cannot explained what happened - all was objectively safe, but my mind felt very uncomfortable. Try it several times, and the sensation will disappear.

Typical height for helicopters is 500-1,500 AGL, and almost all flights are conducted in this regime.
Below 1,500 AGL the surface is quite close and the ground moves by fairly quickly, so you have a good point of reference.
At 3,000 AGL, you can feel quite disconnected from the earth and can have a bout of altitude fear.
When I had to circle up above out of a valley airport in an airplane, with minimum VFR visibility due to smoke, I felt like I had to keep cross-checking my instruments to maintain airspeed and bank angle. It was an uncomfortable feeling, even having had lots of hood time and actual IMC. So its not unusual for unusual flight regimes to have an unsettling effect.
The circling could have also cause some disorientation.
The most important thing, which you accomplished, is recognizing the signs, not panicking, and you kept flying.

That was my first guess a well: not having the ground reference that he was used to. If his spiral was a constant rate turn in the same direction, it could also have been a somatogyral illusion caused by the fluid in the inner ear coming to rest during the turn, and then moving after he returned to level flight leading to a feeling that he was turning even though his eyes told him that he wasn't.  Put an unexperienced, non instrument rated pilot in this situation, in IMC, and it often turns in to a graveyard spiral...